Add "Combination Sum" backtracking algorithm.

Author: trekhleb

README.md

@@ -70,6 +70,7 @@ a set of rules that precisely define a sequence of operations.
   * `A` [Shortest Common Supersequence](src/algorithms/sets/shortest-common-supersequence) (SCS)
   * `A` [Knapsack Problem](src/algorithms/sets/knapsack-problem) - "0/1" and "Unbound" ones
   * `A` [Maximum Subarray](src/algorithms/sets/maximum-subarray) - "Brute Force" and "Dynamic Programming" (Kadane's) versions
+  * `A` [Combination Sum](src/algorithms/sets/combination-sum) - find all combinations that form specific sum
 * **Strings**
   * `A` [Levenshtein Distance](src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences
   * `B` [Hamming Distance](src/algorithms/string/hamming-distance) - number of positions at which the symbols are different
@@ -156,6 +157,7 @@ different path of finding a solution. Normally the DFS traversal of state-space
   * `A` [Hamiltonian Cycle](src/algorithms/graph/hamiltonian-cycle) - Visit every vertex exactly once
   * `A` [N-Queens Problem](src/algorithms/uncategorized/n-queens)
   * `A` [Knight's Tour](src/algorithms/uncategorized/knight-tour)
+  * `A` [Combination Sum](src/algorithms/sets/combination-sum) - find all combinations that form specific sum
 * **Branch & Bound** - remember the lowest-cost solution found at each stage of the backtracking
 search, and use the cost of the lowest-cost solution found so far as a lower bound on the cost of
 a least-cost solution to the problem, in order to discard partial solutions with costs larger than the

src/algorithms/sets/combination-sum/README.md

@@ -0,0 +1,60 @@
+# Combination Sum Problem
+
+Given a **set** of candidate numbers (`candidates`) **(without duplicates)** and 
+a target number (`target`), find all unique combinations in `candidates` where 
+the candidate numbers sums to `target`.
+
+The **same** repeated number may be chosen from `candidates` unlimited number 
+of times.
+
+**Note:**
+
+- All numbers (including `target`) will be positive integers.
+- The solution set must not contain duplicate combinations.
+
+## Examples
+
+```
+Input: candidates = [2,3,6,7], target = 7,
+
+A solution set is:
+[
+  [7],
+  [2,2,3]
+]
+```
+
+```
+Input: candidates = [2,3,5], target = 8,
+
+A solution set is:
+[
+  [2,2,2,2],
+  [2,3,3],
+  [3,5]
+]
+```
+
+## Explanations
+
+Since the problem is to get all the possible results, not the best or the 
+number of result, thus we don’t need to consider DP (dynamic programming),
+backtracking approach using recursion is needed to handle it.
+
+Here is an example of decision tree for the situation when `candidates = [2, 3]` and `target = 6`:
+
+```
+                0
+              /   \
+           +2      +3
+          /   \      \
+       +2       +3    +3
+      /  \     /  \     \
+    +2    ✘   ✘   ✘     ✓
+   /  \
+  ✓    ✘    
+```
+
+## References
+
+- [LeetCode](https://leetcode.com/problems/combination-sum/description/)

src/algorithms/sets/combination-sum/__test__/combinationSum.test.js

@@ -0,0 +1,24 @@
+import combinationSum from '../combinationSum';
+
+describe('combinationSum', () => {
+  it('should find all combinations with specific sum', () => {
+    expect(combinationSum([1], 4)).toEqual([
+      [1, 1, 1, 1],
+    ]);
+
+    expect(combinationSum([2, 3, 6, 7], 7)).toEqual([
+      [2, 2, 3],
+      [7],
+    ]);
+
+    expect(combinationSum([2, 3, 5], 8)).toEqual([
+      [2, 2, 2, 2],
+      [2, 3, 3],
+      [3, 5],
+    ]);
+
+    expect(combinationSum([2, 5], 3)).toEqual([]);
+
+    expect(combinationSum([], 3)).toEqual([]);
+  });
+});

src/algorithms/sets/combination-sum/combinationSum.js

@@ -0,0 +1,65 @@
+/**
+ * @param {number[]} candidates - candidate numbers we're picking from.
+ * @param {number} remainingSum - remaining sum after adding candidates to currentCombination.
+ * @param {number[][]} finalCombinations - resulting list of combinations.
+ * @param {number[]} currentCombination - currently explored candidates.
+ * @param {number} startFrom - index of the candidate to start further exploration from.
+ * @return {number[][]}
+ */
+function combinationSumRecursive(
+  candidates,
+  remainingSum,
+  finalCombinations = [],
+  currentCombination = [],
+  startFrom = 0,
+) {
+  if (remainingSum < 0) {
+    // By adding another candidate we've gone below zero.
+    // This would mean that last candidate was not acceptable.
+    return finalCombinations;
+  }
+
+  if (remainingSum === 0) {
+    // In case if after adding the previous candidate out remaining sum
+    // became zero we need to same current combination since it is one
+    // of the answer we're looking for.
+    finalCombinations.push(currentCombination.slice());
+
+    return finalCombinations;
+  }
+
+  // In case if we haven't reached zero yet let's continue to add all
+  // possible candidates that are left.
+  for (let candidateIndex = startFrom; candidateIndex < candidates.length; candidateIndex += 1) {
+    const currentCandidate = candidates[candidateIndex];
+
+    // Let's try to add another candidate.
+    currentCombination.push(currentCandidate);
+
+    // Explore further option with current candidate being added.
+    combinationSumRecursive(
+      candidates,
+      remainingSum - currentCandidate,
+      finalCombinations,
+      currentCombination,
+      candidateIndex,
+    );
+
+    // BACKTRACKING.
+    // Let's get back, exclude current candidate and try another ones later.
+    currentCombination.pop();
+  }
+
+  return finalCombinations;
+}
+
+/**
+ * Backtracking algorithm of finding all possible combination for specific sum.
+ *
+ * @param {number[]} candidates
+ * @param {number} target
+ * @return {number[][]}
+ */
+export default function combinationSum(candidates, target) {
+  return combinationSumRecursive(candidates, target);
+}